KCSE 2025 Maths P1 Q24 — Differentiation (Maximum Area)

KCSE 2025 Form 4 Calculus

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The Question

“A plot of land ABCDEF has AB = 50 m, DE = 40 m and all angles are right angles. The owner buys an adjacent rectangular plot BCDG with sides x and y so that the whole piece of land becomes the rectangle AGEF. Given that the perimeter of BCDG is 60 m, write y in terms of x, write a simplified expression in x for the area of AGEF, then find the dimensions of BCDG that maximise this area and the maximum possible area.”

1

Write y in terms of x

The added plot BCDG is a rectangle with sides x and y, so its perimeter is twice the sum of the two different sides. Set that equal to 60 metres, divide through by 2, and rearrange to make y the subject. This gives a single expression for y so that everything can later be written in terms of x alone.

2(x+y)=602(x + y) = 60
x+y=30⇒y=30−xx + y = 30 \Rightarrow y = 30 - x
2

Express the area of AGEF in terms of x

The whole plot AGEF is a rectangle. Its width AG is the original 50 m plus the added x, and its height GE is the original 40 m plus y. Substitute y = 30 - x so the height becomes 70 - x, then multiply out the two brackets to get the area as a single quadratic in x.

AG=50+x,GE=40+y=40+(30−x)=70−xAG = 50 + x, \quad GE = 40 + y = 40 + (30 - x) = 70 - x
A=(50+x)(70−x)=3500+20x−x2A = (50 + x)(70 - x) = 3500 + 20x - x^{2}
3

Differentiate and set the derivative to zero

Area is largest at a turning point, where the rate of change of area with respect to x is zero. Differentiate the area expression term by term, set the derivative equal to zero, and solve for x. This locates the value of x that gives the maximum.

dAdx=20−2x\frac{dA}{dx} = 20 - 2x
20−2x=0⇒x=1020 - 2x = 0 \Rightarrow x = 10
4

Find the dimensions of BCDG

Now that x is known, put it back into the earlier expression to find y. These two values are the side lengths of the added rectangle BCDG that make the whole plot as large as possible.

y=30−x=30−10=20y = 30 - x = 30 - 10 = 20
x=10 m,y=20 mx = 10\ \text{m}, \quad y = 20\ \text{m}
5

Calculate the maximum area

Substitute x = 10 back into the area expression for AGEF to find the greatest possible area of the entire plot of land.

A=3500+20(10)−(10)2A = 3500 + 20(10) - (10)^{2}
A=3500+200−100=3600A = 3500 + 200 - 100 = 3600

Final Result

The rectangle BCDG has dimensions 10 m by 20 m, and the maximum possible area of the entire plot AGEF is 3600 square metres.

Why this method works

Area written as a function of x is a downward-opening quadratic, so it has a single highest point. At that highest point the graph momentarily stops rising and starts falling, which means its gradient is zero. Differentiation gives the gradient of the area function, so setting the derivative equal to zero pinpoints exactly the x value at the top of the curve. Because the coefficient of x squared is negative, this stationary point is guaranteed to be a maximum rather than a minimum, so we can be confident the answer really is the largest area.

✓

Since y = 30 - x = 20, height GE = 40 + 20 = 60 and width AG = 60, so AGEF is 60 m by 60 m, giving 3600 square metres, which matches.